00:01
We want to find a plane equation where these three points is on the plane.
00:05
Now before we do that, let's look at some plane forms.
00:10
If i have a plane and i have a point, i name it x1, y1 and z1, and it's perpendicular to a vector.
00:23
Let me call this vector n and it will be abc here.
00:29
Now this n will be the normal vector of the plane.
00:31
Then the standard form of the plane equation will be a times x minus x1 plus b, y minus y1 plus c, z minus z1, equals to 0.
00:49
But we only have three points.
00:52
So how are we going to find the normal of the plane? so it's pretty easy.
00:58
The concept is that if i have three points, a, b, and c.
01:08
All i need to do is to find vector ab and vector ac.
01:13
And i will take the cross product of these two vectors.
01:19
So i will get a vector that's perpendicular to both of them.
01:23
So this is ab cross ac.
01:27
Now, this vector here will be parallel to the normal vector.
01:35
So that's how we can find the normal vector.
01:39
And then we can just use one point here, the x1, y1, z1, and get the plane equation.
01:49
So in this case, let's my 0 -0 be my a, point a.
01:59
My point b will be 1, 2, 3, and my point c will be minus 2, 3.
02:14
Okay, so i will attempt to find vector ab and vector ac.
02:26
So vector ab, you'll be taking the components of point b minus the components of point a.
02:35
So that will be 1 minus 0, 2 minus 0, and 3 minus 0.
02:46
So my ab vector is 1, 2, 3.
02:53
Ac will be taking the endpoint, which is c, minus a in its xyz component...